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Question

# In the Young's double slit experiment, the distance between the slits varies in time as d(t)=d0+a0 sinωt; where d0,ω and a0 are constants. The difference between the largest fringe width and the smallest fringe width, obtained over time, is given as :

A
2λD(d0)(d20a20)
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B
λDd20a0
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C
λDd0+a0
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D
2λDa0(d20a20)
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Solution

## The correct option is D 2λDa0(d20−a20)Fring width, β=λDd For β=βmax, d=dmin And for β=βmin, d=dmax Now, d=d0+a0 sinωt ⇒dmax=d0+a0 and dmin=d0−a0 ∴βmin=λDd0+a0 and ∴βmax=λDd0−a0 βmax−βmin=λDd0−a0−λDd0+a0 =2λDa0d20−a20 Hence, (B) is the correct answer.

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